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(Updated) Why do vertical distances appear much larger than horizontal distances?
Note: I have reformulated this question and provided more concise examples to avoid confusion and allow for better discussion.
According to Google Earth, the distance from where the photo below was taken to the highlighted car is about 450 feet. Which, all things considered, is a short distance (for walking, driving, and flying).

Now, the images below were taken at 500 feet.

Corrected for the height of the flight deck. Credit.

Credit.
It is difficult to grasp the idea that, if I could hypothetically walk downwards from those airplanes to the ground, I would travel the same distance as from where the first photo was taken to the car.
What causes the impression that vertical distances (such as the approach views above) appear much larger than horizontal distances (such as the street above)?
P.S. We find another great example of this illusion in the second photo, where the distance from the beginning of Runway 22 to the aiming point (two white rectangles) is 1000 feet. In other words, that distance is double the height of the airplane in the photo, while it appears to be way higher than that.

For some reason I was staring at the second picture for a long time. It was when I realized where it is. I live in Campinas for almost 15 years now. :)

How cool. I also used to live there.

Can your question be paraphrased:  Why does the runway look longer when viewed from a higher angle like the first photo, but look shorter viewed from a low angle like the second photo?  Because the answer you accepted explains this perfectly, but that didn’t seem to be what you were asking.  My interpretation was:  Why does 500’ up feel higher than the 500’ markers look…”

@Michael Hall: Not really. The runway is just an example. If you look down a 500-feet-long residential road, it doesn’t appear that long. However, if you are, for example, flying 500 feet above a neighborhood, it seems very high up, and not nearly the same distance as the aforementioned street.

OK, but it's the angular effect you are asking about then, not the feeling of height being greater?  For example, if I look at a 12" ruler from 90 degrees off it looks 12", but if I look at it end on, or nearly so, I can't perceive the full length correctly.  Because if that's the case then I grossly misinterpreted your question and will delete my answer.

The question is 100% about the “feeling,” as you described. Not about anything else. In hindsight, I would have phrased the question better and provided more concise examples.

OK, but the answer you accepted is 100% about viewing angle.  (P.S.  There is an edit button...)

It was the best answer, all things considered. (P.S. Do you think it’s still worth it to edit the question?)

If you would like to edit the question to match the answer you accepted I would suggest "Why does the runway look longer when viewed from a higher angle like the first photo, but look shorter viewed from a low angle like the second photo?"  Your call...

@Michael Hall: I have completely redone the question. Hopefully it will help develop a better discussion on the topic.

I undeleted my answer because I think it is relevant...

Please also keep in mind that photos typically are created by a lens that has different properties than the human eyes (e.g. wide-angle or telephoto). So the effects seen on photos are most likely not what you see/saw in reality. So you cannot compare distances from photos made with different lenses.

@jpa: If you are saying the distance to the runway in the last photo is about 1,000 feet, it is not. Using Google Earth, the distance to the threshold is 6,800 feet.

@CharlesNicholson Ah, indeed - I mixed up which description applied to which image!

Have you ever stood underneath a landing aircraft (outside of the airport boundary) and looked up?

The problem is not the elevation; it's the angle you are looking at it.
Looking from top down, you'll see the real proportions, while looking at 30° you'll see half the size (directed towards the line of sight). At 3° it's only a bit more than 5%.
Technically it's the sinus function.
The size perpendicular to the line of sight is influenced by the distance only.
Here is a sketch illustrating the principle in 1D using a line instead of a 2D rectangle (note: I just used the center of the line; actually beginning and end have different angles and distances):

I used three sample points at the same height, but with varying distance, and a sample point at same distance (well: almost), but different altitude.
You'll see that the line will be shorter when reducing the altitude much more than increasing the distance would.
(I know: The sample should have been below the "Distance 92" point for better comparison)
Also not I used rather arbitrary scaling for reasonable numbers to compare: Viewing down at a distance of 100 would yield 100% virtual size.
Alternative Illustration
The following illustration has a black element in the lower right that has equal horizontal (H) and vertical (V) size.
Then there are three reference points:

The blue starting point looking down at 45°
The closer (but same height) red point looking mostly down
The lower (but same horizontal distance) green point looking mostly forward

Colored lines illustrating the center (corner) of the black object, as well as the borders (front and top).

The rest is a bit complicated, so take the time to understand:
The thicker more intensely colored lines attached to at least one corner of the black object are perpendicular on the corresponding view direction, thus a projection of the black object's sizes.
The colored letters H and V mark the corresponding proportions of the projected horizontal and vertical size, respectively.
So for the blue view H and V are equal size, while for the closer red view the vertical element seems much smaller than the horizontal one.
Finally for the lower green view the vertical component of the black element seems much larger than the (projected) horizontal one.
"3D" Illustration
Finally a "3D" illustration of the effect of lowering the altitude shortening the horizontal size (as explained above).
The object looked at is a square with a "center cross", and the relative sizes should be correct (while the absolute size was "zoomed in a bit" to let you see some details):

I hope I got the mathematics right.
The numbers right to each image are:
lateral offset (X),elevation (Y),distance (Z) (negative)@viewing angle (A)
The first three numbers are integers anyway, and the angle is rounded to whole degrees.
The distance is constant the elevation increases (top to bottom, left to right).

From the first photo, it does seem like the shrinking phenomenon is restricted to the dimension parallel to the line of sight, since both perpendicular dimensions (up-down and side-to-side) won't experience this angle-dependent shift. The runway in the photo is 150ft wide, about 1/3 the distance between touchdown markings, but it sure looks like a lot more than that. This explanation accounts for the observation that front-to-back distances shrink, but not side-to-side distances.

@NuclearHoagie That's foreshortening -- you're looking at the runway at an angle, so the "vertical" dimension in your field of view is compressed by geometry, while the horizontal is shrunk only by distance.

This is a psychological illusion, roughly similar to the one that makes the Moon look huge when it's close to the horizon, even though you can still cover it with a fingertip at arm's length.
Essentially, we're used to looking long distances along the ground (if you're standing on the ground, the horizon is around three miles/4.7 km away), and the way distances make objects look smaller -- but few if any of us grew up with enough flight experience to be equally used to looking down similar distances to the ground, or to seeing buildings, cars, runways, and farm fields shrunk by perspective the same way a skyscraper seen from the edge of a city is.

Isn't the horizon more like ~3mi/5km for 1.7m eye height?

Ack, you're much closer to correct, formula says about 4.7 km over sea for a 1.7 m eye height.

Refraction will do the last 300m ;)

Welcome new Max.  ;)  I decided to change my screen name to "Max R" today to disambiguate.

Overestimating vertical scale isn’t limited to aviation, people often misjudge height above the ground.  The reason you “feel” higher than you are is the effect of gravity.  It is not a visual illusion, it is a psychological phenomenon - We are used to being anchored to the surface, and being elevated is an unnatural state so our eyes and brain process this information differently.
Although it is characterized as a “weak force” compared to magnetic and other forces, gravity exerts tremendous influence on our lives.  It severely limits our travel in the vertical and imposes severe consequences for any misstep.  Respect for its effects are ingrained in our DNA - A rational fear of heights is a good survival instinct, but it causes us to subconsciously overestimate actual vertical distance.
Consider the following examples:  We think nothing whatsoever of walking 10’ across a small room.  It takes little physical effort, and exposes us to minimal risk.  However, 10’ up a ladder (not to the top of a 10’ ladder, but your feet 10’ off the ground!) is an entirely different story.
There’s not a lot of effort or risk as long as you are steady, but if you slip and fall gravity will accelerate you to bone shattering speed in the blink of an eye.  You risk significant injury if you don’t land just right.  Some people even fear jumping from 10’ into a soft medium like water, all rational people should fear stepping off their roof to land on their driveway.
Bump the example up tenfold:  100’ is roughly the length of supermarket aisle, while 100’ up is a significant climb up a very tall tree and risks probable death, or at least permanent physical disability as a result of severe injury if you fall.
Another tenfold:  1000’ is short walk down a rural road, a couple blocks of city streets, or the length or a mall.  But 1000’ of vertical is an arduous hike for most, or an extreme climb to get to the top of a major Yosemite sized cliff.  Falling is certain death.
The additional time and effort required to achieve displacement in the vertical, plus the danger element, skews our perception of this axis.
There are no visual illusions causing us to be unable to accurately estimate vertical distance - Consciously we are just as capable of estimating dimensions in the Z axis as any other direction.  Subconsciously, however, our brains perceive vertical distance differently because of the ingrained instinct to avoid falling.

This doesn't really explain the visual difference between looking down 70+ floors to a street and looking three blocks along that same street -- but it's a good answer to the psychological side of the perceived difference.

My point is that there is no actual visual difference, it's a subconscious perception.  (unless you are referring to a horizontal layer of haze or other phenomenon...)

We're saying the same thing, I think.  We're not used to seeing the tops of cars or roofs of houses, but if you chopped down a tree to lie horizontally, or turned a car on its side, it would look very out of place from above, when you're seeing the tops of everything else.  The "everything is smaller than it would be the same horizontal distance away" is a variation of the Moon Illusion, as in my answer.

I agree things look different, but I'm not totally sure we are talking about the same thing...  Our perception of something further away looking smaller is a function of the reduced angles causing the image projected on the retina to be smaller.  It works the same way in all directions.  Consider railroad tracks and a "vanishing point".  Yeah, it would look out of place looking straight up or down a set of tracks, but the geometry is identical to horizontal tracks.  Or am I missing your point?

When you can see the perspective lines (like in a city looking along a street, or looking along a railroad track), your mind compensates for objects looking smaller at a distance.  This is why the Moon looks huge on the horizon -- it's silhouetted behind several treetops, for instance, which you know to be large objects.  When it's high in the sky, it's just a half degree circle with some texture inside -- unless a jetliner crosses its disk.  Similarly, a car three blocks away doesn't look "tiny" because you can see the pespective, but from 500 feet up you can't compensate.

Does my downvoter have any feedback or suggestions to improve the answer?

+1 Interest only: Falling from 100 feet without any impediment along the way is essentially certain death. Some few may survive. V ~= 88 km/h. Deaths ~= kph^2/5000 --> so at 88 kph death probability ~=  150%. The velocity does not allow for drag but will not be vastly slower than the above.

@RussellMcMahon, you are probably right, I was just looking for some sort of progression in the examples…

The perception of human vision is angular rather than linear.  What our eye physically sees is an object that is 3 degrees wide.  Our brains then use an assumption about the object's size to to estimate how far away it is, OR uses an assumption about how far away the object is in order to estimate its size.
A 747 10 miles away takes up as many nerve endings on your retina as a Cessna 172 that is 1 mile away.  You look at the shape of the 747, you recognize it, and your brain is able to "reverse engineer" that it's a long ways away.
If you are standing a mile away from a runway, that runway still only "subtends" or "spans" a few degrees across your retina.  That sends a powerful signal to your brain, "that's small."
If you are standing a mile away from an antenna tower the same length, it will literally subtend 45 degrees or more of your retina and sends a powerful signal to your brain, "that's big."

This explains why things look smaller at a distance, but doesn't address any difference in perceiving distance vertically vs horizontally.

This also explains why large planes seem to fly slower than small planes. Our brains can reverse engineer the size, but the angular speed is just too much to compute correctly.

A large psychological factor here is foreshadowed by Zeiss Ikon.  Our perception of distance using focus or parallax is only effective up to about 10m from our eyes.  A round dot 50m away looks pretty much like a larger round dot 50km away.
The moon looks larger on the horizon because there's reference points...maybe a city skyline, or some trees, but enough to see that, compared with these familiar references of known size, the moon is both larger and more distant.  When well up in the sky, there's no visceral evidence to indicate definitively that it's much more than 10m away.
Similarly, your first picture shows sub-markings, access roads, other aircraft, scenery, and (I think) a train to provide size and distance references both beyond and between the marks.  In the second, however, there is nothing between the plane window and the ground (which is in fact over 500ft distant, since the camera is not perfectly vertical).  The mind registers that as >10m, and doesn't try too hard to improve the estimate, since height information is interpreted with heavy weight on "how bad would it be if I fell from here" and 10m isn't much different than 1km in that equation.


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What is "taxi speed" for an Airbus?
In the FCOM we can see a lot of "taxi speed", for example after landing:

AT TAXI SPEED, ALL REVERSER LEVERS......STOW

But what is taxi speed? Is there any definition or particular number? 40kt or 30kt?

The Flight Crew Training Manual (FCTM) defines taxi speed as a ground speed of up to 30 kt:

Taxi Speed and Braking
On long, straight taxiways, and with no ATC or other ground traffic constraints, the PF
should allow the aircraft to accelerate to 30 kt, and should then use one smooth brake
application to decelerate to 10 kt. The PF should not ”ride” the brakes. The GS indication on the ND should be used to assess taxi speed.

(Airbus A320 FCTM - Normal Operations - Taxi)

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

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Can the flap on one wing be used to reduce a left turning tendency?
I fly a single engine Jabiru UL45 and it has a tendency to turn left during straight and level flight.
Is it possible, in order to reduce a "Heavy Wing" to slightly increase the degree of the flap on that wing?

So probably no aileron trim?  Is the ball centered?  Does it roll left at cruise power?  Is the wing fuel balanced?

You’re talking about a very small airplane whose CG placement left/right will be strongly affected by you being in the pilot seat.  Have you flown with a similar-sized passenger, or put some ballast sandbags in the right seat?  Would be interesting to know if it’s aerodynamic/rigging, or just weight distribution that is causing this.

This is easily done if your plane has trim tabs on the wing. This is a small strip of metal attached to the trailing edge of the wing which you bend with pliers little by little on the ground until the plane flies straight and level with the aileron trim centered.

If you can rig the flaps at the retracted position to a slight differential, while staying within the rigging tolerances yes you can tweak the flaps to correct a rolling tendency. So if flaps zero is zero +/- .5 degree, there's nothing stopping you from adjusting the right flap up to .5 degree up, and left up to .5 degree down, to correct a left wing heaviness.
This is done on commercial aircraft all the time. Provided you stay within the published tolerances.

No, probably not.  Most aircraft are designed to deploy flaps symmetrically.  If your aircraft has aileron trim, use that.

Unfortunately Michael it does not have aileron trim

Then how about rudder trim?

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What is this nose projection on the Tupolev Tu-95?
The Tupolev Tu-95 has this long and quite substantial appearing projection from it's nose just under the cockpit windscreen:

What is this? I initially though this might be a cannon but being a heavy bomber, that makes little sense as it would be quite difficult to aim effectively.

I always assumed it is an air-to-air refuelling probe.

That boom is the refueling probe used for in-air refueling:

(image source: pinterest)
You can see it in action in this photo, where an Il-76 is refueling a Tu-95:

(image source: reddit)


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Does an orbital trajectory reduce my required lift for high mach numbers?
when an aircraft is flying from let us say Paris to New York on a fixed altitude it is not flying on a straight line but on an orbital / circular trajectory around the earth. Assuming, the aircraft is flying Mach 5 or even faster, does this significantly reduce the amount of required lift, since the centrifugal force is acting on the aircraft?

We can calculate this by comparing the centrifugal force to the force of gravity:
$$ F_\text{centrifugal} = m \omega^2 r $$
$$ F_\text{gravity} = G \frac{m M}{r^2} $$
The ratio is then given by
$$ \frac{F_\text{centrifugal}}{F_\text{gravity}} = \frac{m \omega^2 r}{G m M / r^2} = \frac{\omega^2 r^3}{G M} $$
Let us assume we fly at FL600 (where Concorde flew), so $ h \approx 18 \, 288 \, \text{m} $ and then $ r = R + h$ (with Earth radius $R$ and mass $M$). The speed of sound at FL600 is $ c \approx 295.1 \, \text{m/s} $, which gives a speed at Mach 5 of $ v = 5c \approx 1476 \, \text{m/s} $. On a circular "orbit", we have $\omega = v/r $. Plugging in the numbers, I get
$$ \frac{F_\text{centrifugal}}{F_\text{gravity}} \approx 3.48 \, \% $$
So flying at Mach 5 would reduce the required lift by about 3.5%.
However, as Ralph pointed out in the comments, you need to take the Earth's own rotation into account. If you are on the equator, the above effect is already ~0.3% without moving at all (w.r.t. the Earth's surface). You need to add or subtract the Earth's motion from the velocity above. On the equator, that gives a difference of about 463 m/s (add when moving East, subtract when moving West). Repeating the calculation above then gives:
$$ \text{East:} \; \frac{F_\text{centrifugal}}{F_\text{gravity}} \approx 6.02 \, \% \qquad \text{West:} \; \frac{F_\text{centrifugal}}{F_\text{gravity}} \approx 1.64 \, \% $$

Is this a case where the rotational velocity of the Earth would actually affect the numbers, i.e. adding it to an eastbound jet, subtracting it for a westbound, and (roughly) no change for a north or southbound jet?

@RalphJ Good point! In space, there is no difference between retrograde and prograde orbits (Space.SE). But in the atmosphere, it can make quite a difference. I added a paragraph with some more details.

So the tl;dr might be something like "Yes, but not much, and mostly if you're flying towards the East.  The accountants will be pleased at a bit of fuel saved."

@WayneConrad: The extra lift scales with speed squared, so Mach 10 would have 4x the effect, about 14% +- east / west difference.  But yeah I guess that's still "not much", and much higher speeds Mach 20 is more like re-entry from actual orbit.  Even Mach 10 is much less realistic than Mach 5.

@WayneConrad I think the accountants will be pretty mad at you for flying at Mach 5 because that really increases fuel consumption :D


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Why does secondary surveillance radar use a different antenna design than primary radar?
In radar surveillance we use primary and secondary surveillance radar for different purposes.
I'm wondering about the antenna that use in both system:

Secondary Radar uses an array antenna
Primary Radar uses a reflector

Why the different designs?

With the secondary system, you are simply sending a one-way message from the radio tower to the transponder in the plane, and the transponder is transmitting a message back from the plane to the radio tower.  It is a very simple two-way radio conversation between relatively high powered stations.
With primary radar I am sending out a signal and all I am receiving back is the nearly microscopic amount of energy that bounced off your aircraft.  That signal may be hundreds of thousands--or even millions--of times weaker than the secondary signal being transmitted by your transponder.
From an engineering perspective, the primary receiver is designed to solve a far, far more difficult task, hence the more complicated antenna requirement.

If I may extend your answer from a physics perspective: the primary radar is (usually if not always) a different frequency (and thus wavelength) from secondary radar. For a certain design, antenna size is typically a factor of the wavelength. With two different frequencies, a different antenna design may have been necessary just to get the sizes in the right ballpark. With different sensitivity requirements, a less sensitive SSR antenna design that allows it to be small is likely to be selected.


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What is the definition of load factor and how do you apply it?
I'm a little confused about the definition of the load factor, n, as the definitions I found seem to contradict its application in force and moment balance analysis of an aircraft structure.
The definition given in my university script states that the load factor equals the ratio of all external forces acting on an aircraft minus its weight, to the magnitude of the aircraft's weight; this implies that the load factor is the ratio of all aerodynamic forces acting on an aircraft to the magnitude of its weight.
The Wikipedia definition of the load factor states that the load factor is the ratio of the sum total of all aerodynamic forces acting on an aircraft to its weight.
My script definition therefore tallies with the Wikipedia definition. All good so far.
Now, however, there is also a sentence in my script that states that the load factor is a means of including all inertial loads and gravity loads in observing the loading of an aircraft; this implies that weight is included in observing the loading of an aircraft: this makes better sense to me, since, when expressing the loading of an aircraft, it would be easier to say that the aircraft is now being loaded "this many times its weight", compared to this many newtons; this however, contradicts the definitions above.
And now to the force and moment balance analysis issue, I noticed that all components of an aircraft are expressed in terms of its load factor, which is fine, but then in performing the balance, the weight of the components are not considered. How can you ignore the weight force which is always acting on the aircraft when the load factor (per definition) excludes it? An added confusion is in that lift is considered in the moment balance, with all component's loading expressed in terms of load factor; how can you include lift, when it is already incorporated into the load factor of the components; this implies that the part is further loaded above its current loading.
I would appreciate if someone can straighten up what the load factor means, and how should it be applied in force and moment balance calculations.

I thought "load factor" was a business economics term for the percentage of revenue seats filled on a commercial aircraft (for a given flight).  The load factors of multiple iterations of the same numbered flight are averaged, as in "Southwest Airlines' flight WN4604 has an average load factor of 89%".  Perhaps I'm confusing this metric with another name...

Guha, weight is not a "real" force, it is an artificial force, like centripetal force, used to make the math work because we are doing our calculation on the earth, in an accelerated frame of reference. Weight does not act on the aircraft. Only aerodynamic forces do. But since we are in an accelerated frame of reference, the "No force" trajectory is opposite the acceleration vector of entire frame of reference. i.e, a ballistic arc constantly curving downwards - away from the acceleration vector (which is upwards.)

I looked at the definition of "load factor" on Wikipedia. They show 4 or 5 meanings for the term, two of which are associated with aviation, so you might consider changing the title of the question to reflect which one you want.

The load factor of an aircraft describes the mass loading of an aircraft as a multiple of its weight. The same definition applies if we are speaking about an aircraft component: the load factor, here called the local load factor, will be the mass loading of that component as a multiple of that component's weight.
First we must understand what mass loading is. The mass load of an aircraft is the inertial loads, as a result of the acceleration of the aircraft and the weight of the aircraft, as a result of the acceleration of gravity. Since the inertial loads always act opposite to the aircraft's acceleration, it is vectorially signed negative. Gravity, as it always points downwards, is in sync with our coordinate system which defines the downward direction as positive; hence, it will be vectorially signed positive. Summing the two vectors will give us the total mass loading of the aircraft. Dividing this total loading by the aircraft's own weight will give us an idea as to how much more that its own weight is the aircraft is currently being loaded by the maneuver it is performing, and by that, how structurally strong it is or needs to be.

Really boils down to load forces created by acceleration and gravity, separately considered, then applied to the orientation in the gravity field.  Your answer provides a starting point by using the universal and unchanging gravity vector as a standard reference signed positive.

The load factor is the total acceleration you feel, pointing downwards. In straight and level flight, the load factor is 1: you only feel the acceleration of gravity. So a load factor 1 equates to 9.81 m/s$^2$ (If gravity was higher, let's say 15 m/s$^2$. load factor 1 would equate to 15 m/s$^2$. But that's another story.)
Image source
Bank the aircraft 60° and fly a co-ordinated turn, and you'll experience a downward acceleration of 2g. This is a case that is easy to understand since it is a static situation with constant velocities. If we have a dynamic sine wave for instance, the load factor would be a function of where we are in the cycle, if the aircraft is accelerating up or down. The actual acceleration is added to the gravity vector.
Load factor does not ignore the weight of the aircraft:

In steady horizontal flight, load factor is 1 because it is the same weight that would show up on a scale on the surface of the earth.
All dynamic accelerations are added to 1. If the aircraft accelerates downward with -1g, the load factor is zero and everyone on board is weightless.
Forces are vector entities. For the load vector, we're only counting the resulting accelerations in the z-axis of the aircraft. After summing the forces in 3 degrees of freedom, the z-axis force F drives the airplane up/downwards with an acceleration of a = F/m.

Yes, but can you provide an explanation from a structural point of view. Thank you.

Load factor is simply Aerodynamic Lift divided by aircraft weight, or more correctly, by aircraft inertial mass as normalized by 1 "G" - (32.2 ft/sec2 or 9.98 m/sec2. - CORRECTION (thx to @Michael Kjorling : 9.80665 m/sec2)
The units are dimensionless (if both Lift and weight are measured in the same units), but we commonly refer to the load factors in "G"s.  It really doesn't matter what the angle of bank is. Even if you are upside down, if the wings are producing twice as much lift as the aircraft weighs, you have 2G's Load factor.
To address the unnecessarily complicating issue of gravity, you must understand that considering gravity is only necessary to determine the motion or flight path because an aircraft in flight is moving in an accelerated frame of reference. It would be exactly the same as if your aircraft was in an enormous closed box, filled with air, in outer space, under acceleration from a giant rocket strapped on one side, that accelerated the entire box at 32.2 ft/sec2.
To hold a constant "altitude" in the accelerated frame of reference of this giant box, you would have to trim the aircraft so that the wings were producing lift equal to the "weight" (mass x 32.2ft/sec2) of the aircraft. The Lift on the wings (if it equals the weight, just produces an "upwards" acceleration of 32 ft/sec2, which matches the 32.2ft/sec2 that the box is accelerating, and keeps the aircraft the same distance above the "floor" of the box.
On the earth, we are in an identical 32.2 ft/sec2 accelerated frame of reference, except that is being caused by the gravitational field, and not by an external rocket.

Do you have a citation for the number 9.98 m/s², or can you at least clarify where it came from? That's quite far from the 9.81-9.82 m/s² commonly used as an approximation for Earth's gravitational accelleration.

Ahhh, I was just remembering the value from when I was in undergraduate school. As it's been quite a few years (1967-1971) since then I am actually surprised I was as close as I was. But you are right. Wikipedia says (https://en.wikipedia.org/wiki/Standard_gravity) 9.80665 m/sec2.

The best answer above is from Charles Bretana But, additionally can be calculated from : 1/cos(bank angle)

User31584.., your answer is only valid when in level flight, i.e., when maintaining constant altitude. It is an a simplification. Very useful as it makes calcukation of load factor simpler, (under thst special condition), but True only when in that special condition.

If you are flying straight and level, the weight of your plane is in equilibrium with the lift, and the load factor is lift/weight = 1.
Imagine that you are flying s/l and –for some strange reason– the intensity of the earth's gravitational field, (usually expressed as the acceleration of gravity, g) suddenly becomes 3g. In order to keep flying straight and level, you will have to increase your airspeed, or to (prudently) increase the AoA of the wings. Under those conditions, your wings will be stressed to 3g. The 'load factor' will rise to 3g.
As we all know, the intensity of the gravitational field is quite stable, and you'll never find yourself flying in the conditions mentioned above, but if your plane follows a curved trajectory, the apparent weight of your plane will be increased by the said inertial forces. If the acceleration (normal to the relative wind) associated with those inertial forces is, for instance, 2g, your wings will be stressed by the load due to gravity plus 2g...

Your definition of load factor here is not quite correct.  If the strength of the gravitational field increases, an aircraft will still have a load factor of 1 in unaccelerated, straight and level flight.  An airplane flying straight and level in Jupiter’s atmosphere with experience a gravitational acceleration 2.4 times that of an airplane flying straight ald level in Earth’s atmosphere, but both aircraft are subject to a load factor of 1.

@Carlo Felicione  To a first approximation, 'inertial forces', i.e. those derived from an acceleration, are indistinguishable from gravitational forces. Hence, a pilot flying s/l can't tell the difference between an increment in the gravitational field and an 'inertial force'.

@Carlo, if you define load factor by Lift divided by weight as defined by the gravity field you happen to be in (using your Jupiter analogy) then yes, but this is really stupid, and would yield silly answers. Using this definition, if you were flying a 1000 kilogram aircraft in a giant box in outer space, filled with air, and this box is un-accelerated, an you pulled back on the stick hard enough to generate 1000 newtons of wing Lift, pushing you into the seat with a force of 5 G's, by your definition, (Since aircraft weight is zero in this box), the load factor would be zero?

No, Load factor should be defined as Lift divided by aircraft inertial mass (normalized) and inertial mass does not change no matter what planet you happen to be on.

...and Romeo, NO, the G-meter on the aircraft on Jupiter would read 2.4 Gs. To keep the aircraft in level flight in a 2.4 G accelerated frame of reference, the wing would have to be generating enough lift equivilent to 2.4 x the aircraft's inertial mass.

As another answer stated, Load Factor is simply Aerodynamic Lift divided by aircraft weight.  Note that if Aerodynamic Lift is known and weight is known, then one does NOT need to know whether or not the aircraft is accelerating to calculate the Load Factor.  If Aerodynamic Lift is known and weight is known, then inertial loads do not need to be considered.  If aircraft X weighs 5000 pounds and the aircraft is generating 5000 pounds of lift, then the load factor is 1.  We can also call the load factor the G-loading-- or at least the component of the G-loading that we measure on a G-meter-- the component that acts in the up-and-down direction in the aircraft's reference frame.  Anyway, if aircraft X weighs 5000 pounds and the aircraft is generating 5000 pounds of lift, then the load factor is 1, regardless of whether the aircraft is in straight-and-level flight, or is banked 45 degrees, or is inverted at the top of a loop.  The aircraft's acceleration through space will be different in all these cases, and in all but the first case the flight path will be curving (accelerating) earthward, but the load factor will be the same.
We can also say that the load factor is the "felt" acceleration, or at least the component of the "felt" acceleration that acts in the up-and-down direction of the aircraft's reference frame.  The total acceleration is equal to the sum of the "felt" acceleration plus the 1-G downward pull of gravity.  Examples-- straight and level flight-- total acceleration 0 G, aerodynamic lift force = 1G * weight, "felt" acceleration 1 G, load factor = 1.   Aircraft inverted at top of loop with 1 G showing on the G-meter-- total acceleration 2 G downward, aerodynamic lift force = 1 G * weight, "felt" acceleration 1 G, load factor = 1.  So if we DO know the "felt" acceleration, or if we DO know the the total acceleration and the aircraft's attitude in space, then we can calculate the Load Factor without knowing the actual lift force in pounds that the wing is generating, and without knowing the aircraft weight.
Knowing the load factor alone does not tell us how much stress is being exerted on, say, the connection between the wing and the fuselage.  This will be affected by how the mass of the aircraft is distributed.  It also will be affected by whether or not the horizontal tail is generating a downward lift force, which requires the wing to generate more lift to achieve a given Load Factor.  For more on this see this answer to the related question "How does an aircraft's weight affect the V-n diagram?"  How does an aircraft's weight affect the V-n diagram?

(Accidental duplicate)

Second paragraph, "Examples-- straight and level flight-- total acceleration 0 G, aerodynamic lift force = 1G * weight" would better read "Examples-- straight and level flight-- total acceleration 0 G, aerodynamic lift force = 1 * weight".  Etc.

This is correct for the load factor at the center of gravity only. Now consider a long fuselage with pilots sitting up in front. Pitch motion and fuselage bending will add their own contribution to the load factor at the pilot position, which may be significantly different from what is measured at the center of gravity.

Thanks @Peter Kampf!  Again a good example of non convention of terms turning physics into witchcraft.  I would try to explain that applying elevator would create stress loads no matter what the net sum Gs were.  Secondly, considering the case of perceived Gs, and applying this aerodynamic stress may be erroneous.  Tried to show horizontal loop is constant aerodynamic Gs with constant speed, Vertical loop of constant speed would involve same aerodynamic loads, but gravity vector at top and bottom are reversed WRT pilot, creating different perceived Gs.

Peter Kampf, roger that re CG only; I guess my answers assume the pilot and G-meter are located at the CG; yes clearly pitch rotation rate will affect "felt" acceleration and G-loading at a station well ahead of or aft of the CG.

As on many threads, this can degenerate into opposing and equally non yielding points of view.  Also stress loading is a very complex subject.  If we can work to agree on terms and definitions, some real progress can be made.

Now, @quiet flyer,  yes, supporting weight on the wing can be considered as a load = 1, and I would be happy to work in those parameters.  This would be, in a practical sense, the spanwise bending stress.  The only issue I see is "aircraft inverted at top of loop--total acceleration 2G downward".  I can see what you are trying to say, and this is why it seems they are trying to separate the gravity component.  So, WRT pilot, Gravity is -1, aerodynamic is +2, total is +1 in the loop. Let's all "stay in the loop"  Bob

I will not down vote your answer, but total 2G downward acceleration will produce total 2G load upwards, the mythical "centripetal" force. So, again, issues of convention.  The concept of "felt" Gs equating to wing spar loading has merit, so I will continue to value its use.

Okay.
A lot of good answers up there.
After going through all of them there was one question that still lingered in my mind. Why do we neglect the vertical component of lift in considering the load factor? Also why should the weight be considered since it is anyway getting offset by the vertical component of the lift force.
The answer to that question simply lies in the way we define load factor. Examining the definition of load factor carefully we understand that it is the net force that the wings of an aircraft have to provide to accomplish a certain kind of maneuver. Since the wings provide this amount of force they experience an equal amount of loading too and this is what we refer to as the load factor. In simpler terms the amount of force that the wings are supposed to provide is simply the amount of force they end up enduring. Consider the same in a coordinated turn. Keep the lift forces aside and examine the other forces that are required for the maneuver along with the weight (which is something that the aircraft structure anyway has to support). This includes the centripetal force and the weight for a simple banked turn of an aircraft. Thus the net force required for the turn is the vector sum of the weight and the centripetal force. This (by drawing the free body diagram of the aircraft) simply comes out to be the net lift force acting on the aircraft. Divide it by the weight of the aircraft and you get the load factor. If the bank angle is $\theta$ then the load factor comes out to be sec$\theta$. And that explains the graph tending to infinity as the bank angle approaches 90 degrees.

We neglect the vertical component of lift because the orientation of the lift has nothing to do with the force of that lift. Load Factor is just a representation of the magnitude of the force the lift is exerting on the aircraft. What direction the force is acting in is not relevant. The aircraft structure does not have to support the weight. It only has to support the force of the aerodynamic force exerted on it, i.e., by the Lift.

If you unload the aircraft (push the stick forward until the G meter reads zero), there is no load on the airframe and the load factor is zero. The weight has certainly not disappeared. The aircraft still weighs what it weighs. But the Lift is gone, so the load is zero.

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text:
Are there any regulations on who can sit next to emergency exits?
Whenever I fly with Lufthansa, if I want to sit next to the emergency exits, I have to satisfy some criteria such as:

Be able to communicate in German or English
Be at least 16 years old
Be able in good physical shape so as to be able to open the door(I don't recall the exact wording Lufthansa used)

Now this is not a full list, but it gives you an idea.
Later on I took a flight with another company (still an European company). This company was proposing to get those seats for an additional cost (as they have much more leg space), and with no criteria to satisfy. Upon boarding the plane, I noticed that the people sitting in those seats were above 60 y/o, so I doubt they would be able to open the door. So I wonder are there any regulations in EU?
I saw an answer for the FAA regulations over here: link, but I wonder how is it in EU.

The requirements vary, but I've never seen a maximum age listed, only the general criteria that they be able to perform the functions required (for window exits, some airlines say you must be able to lift 20kg or similar, as the exit is heavy). Certainly many people above the age of 60 qualify, while a number of people below 60 will not.

EU No 965/2012 provides some direction.

CAT.OP.MPA.165 Passenger seating
The operator shall establish procedures to ensure that passengers are seated where, in the event that an emergency evacuation is required, they are able to assist and not hinder evacuation of the aircraft.

Further clarification can be found from Acceptable Means of Compliance
(AMC) and Guidance Material (GM) to Part-CAT.

AMC2 CAT.OP.MPA.165 Passenger seating
ACCESS TO EMERGENCY EXITS
The following categories of passengers are among those who should not be allocated to, or directed to, seats that permit direct access to emergency exits:
(a) passengers suffering from obvious physical or mental disability to the extent that they would have difficulty in moving quickly if asked to do so;
(b) passengers who are either substantially blind or substantially deaf to the extent that they might not readily assimilate printed or verbal instructions given;
(c) passengers who because of age or sickness are so frail that they have difficulty in moving quickly;
(d) passengers who are so obese that they would have difficulty in moving quickly or reaching and passing through the adjacent emergency exit;
(e) children (whether accompanied or not) and infants;
(f) deportees, inadmissible passengers or persons in custody; and
(g) passengers with animals.

Whether or not a person can sit in the emergency exit row may also depend on if the specific seat is designated as an emergency exit. For example, on narrow body aircraft, generally all seats in the emergency exit row are considered emergency exit seats and only persons who "able to assist" can sit in those locations. However, on some wide body aircraft, only seats on the window sides are considered emergency exit seats. On an aircraft with a 2-3-2 configuration, the 3 seats in the middle are not considered exit row seats and therefore do not have any further safety requirements.

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text:
How do I execute this instrument approach at Larnaca (LCLK)?
I am on course 200 degrees and got a clearance for the VOR-ILS X RWY 22 approach at Larnaca International Airport (LCLK):

Where is the/an IAF in this case? Do I have to join a holding pattern over the LCA VOR for a course reversal to course 041 inbound to the VOR, and then another base turn for a course reversal to the final approach course?

It would be helpful to know where you are (position and altitude) when you receive the clearance.

Let's say  20 nm out of LCA at 6000ft

Unfortunately I can't find LCLK charts anywhere online. But, would you actually be cleared for that approach from that direction? That is, isn't there another, more direct, option?

If you're course 200 then you are more than 30 degrees off the initial outbound track of the approach, therefore you should join the hold prior to commencing.
In more practical terms, if you're VMC at any point and your happy with your terrain clearance, then there would be nothing to preclude a request for a straight in, although from experience, LCA can get busy!


text:
Does the tailwind component always have the same absolute value as the headwind component?
If we are on course 360 and we have crosswind from 020 at 20 kts...does the amount of head wind on this leg equal the amount of tailwind on the reciprocal course (i.e. 180)?
To amplify the question with an example:

TAS:   260kts
Wind velocity: 310/35 kts

Given the above information, calculate ground speed on courses 240T and 060T.
When I use e6b+ aviation calculators, I get the following answers:

On a course of 240, the ground speed is 246kts. This is 14kts slower than the true airspeed.
On a course of 060, the ground speed is 270kts. This is 10kts faster
than the true airspeed.

Since I believe the headwind and tailwind components should be equal, the question is: why is there a difference of 4kts between the computed headwind component and the computed tailwind component

Welcome to the site @Joey. Your question is not clear, what are you asking?

@GdD I think the question is clear, what's not clear is where Joey needs help.

You fly in a mass of air, and that mass is moving 240° SW at 20Kts. If you fly straight north 360° one hour, you'll end roughly 20 miles SW of your theorical destination with no winds. Same applies for a 180° straight south flight : you'll end somewhere 20 miles SW, because your aircraft is drifting within the mass of air. Accuracy is another matter : Your aircraft has a mass, a velocity, is not as "driftable" as a balloon, air being a compressible fluid, and you won't always have 20kts everywhere... You can estimate your ideal course, but you'll have to use references to correct your course.

YES, If the "amount of headwind" was 10 knots due to the crosswind, then the "amount of tailwind" would also be 10 knots on the reciprocal course.
BUT, because the 10 knot disadvantage with a head wind will last for a longer period of time than the 10 knot advantage with a tailwind, the entire trip will take longer and use more fuel than a trip in still air.
EXAMPLE: The headwind leg is 15 minutes longer, but the tailwind leg is only 10 minutes shorter, so you have a net loss of time.

Your answer explains why it will take longer with a headwind that with a tailwind, but does not answer the question of why the OP's computed components are not equal (there is a difference of 4kts).

My original answer was correct for a wind of 020 at 20 before the OP substantially edited his question.  I don't know why his E6B calculations are not equal. My guess is that the increased wind correction angle needed to maintain course effects the ground speed in his 2nd example.

Does the amount of head wind on this leg equal the amount of tailwind
  on reciprocal course ie 180?

Yes, the intensity of the tailwind will be the same than the intensity of the headwind.

In both case the intensity of the head/tail wind component will be
$\left\| W_l \right\| = \left\| W \times cos\; W_a \right\|$
where $W$ is the wind intensity, $W_a$ is the angle between heading and wind. For a 20 kt wind at 60° ($cos\; 60° = 0.5$), the result is 10 kt.
The cross-wind component is also reversed with the same intensity.

Thanks for your reply.this is what i had calculated but when we use aviation calculators like e6b+ to solve the same problem the effective head wind and tail wind component is coming out to be different..

My answer comes from the point of view of a cyclist (also being subject to the effects of headwinds and tailwinds).
I would suggest that any form of consistent winds on a round trip can only act to make the net effort required greater than in calm wind conditions.
Consider the effect of two special cases:

Consider a strong cross-wind, blowing at 90 degrees to the path being travelled (say, a westerly wind). In both directions, (north and south), you will face increased head winds. In an aircraft, you would need to aim somewhat west of your intended direction in order to stay on course. Both legs of the journey would be more difficult.
Consider a strong head-wind, blowing parallel to your course (head-wind in one direction and tail-wind on the way home). For the sake easy maths, let's assume that the wind-speed is half that of the cruising airspeed of your aircraft. Going into the wind, your ground speed will be only half that of still conditions and so the journey will take twice at long. A one-hour flight (in still conditions) would take two hours. To make up all that lost time on the way home you would need to be moving very fast. However the tail-wind would only make your ground speed 1.5 times your normal speed and you would return along the 1-hour path in only 40 minutes to give a round trip time of 2hrs 40mins.

In both cases, and therefore all variations of wind directions and for all wind strengths above 'calm', your total round-trip time will always be longer.
If you are only concerned with differences in airspeed (and not how long it will take to cover a certain distance), then a direct tail wind will help in one direction and a direct head wind will hurt in the other direct with commensurate changes to net ground speed. However, as soon as there is any cross-wind component, that will hurt in both directions, hence the 4kt discrepancy mentioned in the question.

if you have perfect sidewind, I don't see how you can have increased headwind. also, while vaguely related, it does not directly address the question.

Because with a side wind, to maintain the intended course, you would need to head slightly into the wind (and the stronger the wind, the greater the deviation) and consequently for a given airspeed, you would cover less ground in the intended direction.

@David0337 I concur, your effort in correcting is wasted as far as desired direction is concerned, with a head or tail wind.  Though you gain in speed with a tail wind you would gain even more if ONLY the forward component of the same wind had been directly behind you.

How did you come up with 40 minutes rather than 30?

@Steve V My assumption was that the wind speed was half that of the aircraft's airspeed.  Let's assume a 60km distance at 60km/h. In still conditions this would take 1hr (t=s/v). With a 30km/h headwind it would be 60/(60-30) = 2hrs. With a 30km/h tailwind, it would be 60/(60+30)=2/3hr=40mins. To do the run in 30 minutes would suggest a 60km/h tail wind but that would make the upwind leg rather more difficult.

You can replicate the results from the E6B calculation with simple triangles to do the vector addition. https://www.triangle-calculator.com/?what=&q=a%3D260+A%3D110+b%3D35&submit=Solve (and then swap the A=110 with A=70 for the return leg).

@David0337 - Oooops, you're exactly right. That'll teach me to try to do E6-b math in my head rather than going and getting the actual E6-b.


text:
Why do some Canadian Class E airspaces have no ATC frequency on VFR maps?
PPL student here, studying airspace. Looking at the airspace around Vancouver, BC (see Vancouver VTA) I see that most of the lower mainland is class E controlled airspace. "No Contact" tells me that there is no ATC unit to ask for permission. Why is it class E then in the first place? Why isn't it class G?

You'll also find a "No Contact" Class E zone just to the northeast of your arrow, sandwiched in between the Class C TWR (SFC-1900'), and the Class C VR TML (2500'-12,500'). Between 1900-2500 feet, you're in Class E airspace, in which (as a VFR pilot) you're not required to be in contact with ATC.  I'm not familiar with Can VNCs, so I won't submit this as an answer, but I suspect that is what "no contact" means... "contact with ATC not required".  Your implied assumption that "no ATC contact for VFR pilots" equals "always Class G" is what is tripping you up, I think.

In class E airspace, ATC service is provided to IFR flights, but not to VFR flights. The "NO CONTACT" note most likely refers to VFR flights, since you are looking at a chart designed for VFR flight planning. The airspace is class E, and not G, because whoever designed it wanted IFR flights to be controlled within in.
See this related answer for a brief description of the different airspace classes.


text:
Is vertical separation based on geometric difference or altimeter difference?
In the comments below one of the answers to this question about the influence of temperature on vertical separation, the question was raised whether ATC ensure 1000ft vertical separation based on geometric difference (true altitude) or pressure altitude difference.
In cold air, the true vertical separation will be less than 1000ft if the difference in altitude indicated on the altimeter is 1000ft.

Source: the other question
Is there an offical reference that gives the answer?

Air Traffic Control provides vertical separation based on pressure altitude.
ICAO Document 4444 (Procedures of Air Navigation Services - Air Traffic Management) section 5.3 defines vertical separation.

5.3 VERTICAL SEPARATION
5.3.1 Vertical separation application
Vertical separation is obtained by requiring aircraft using prescribed altimeter setting procedures to operate at different levels expressed in terms of flight levels or altitudes in accordance with the provisions in Chapter 4, Section 4.10.
The vertical separation minimum (VSM) shall be:
a) a nominal 300 m (1 000 ft) below FL 290 and a nominal 600 m (2 000 ft) at or above this level, except as
provided for in b) below; and
b) within designated airspace, subject to a regional air navigation agreement: a nominal 300 m (1 000 ft) below
FL 410 or a higher level where so prescribed for use under specified conditions, and a nominal 600 m (2 000 ft)
at or above this level.
Note.— Guidance material relating to vertical separation is contained in the Manual on a 300 m (1 000 ft) Vertical Separation Minimum Between FL 290 and FL 410 Inclusive (Doc 9574).

This explains that the separation is based on altimeter difference of 1000ft. It does not mention low temperature correction when it comes to vertical separation.
Low temperature correction is applied when it comes to the minimum vectoring altitude. Of course the height of obstacles does not change with low temperature so here a correction is needed to avoid collisions risk.

8.6.5.2 When vectoring an IFR flight and when giving an IFR flight a direct routing which takes the aircraft off an
ATS route, the controller shall issue clearances such that the prescribed obstacle clearance will exist at all times until the
aircraft reaches the point where the pilot will resume own navigation. When necessary, the relevant minimum vectoring
altitude shall include a correction for low temperature effect.


text:
Fly over or fly by question
As far as I know, we're to fly-over a RNAV fix if such is drawn surrounded by a circle. But, in non RNAV procedures, how do I know that I must fly-over a fix?
E.g.:Taking this chart into account:

For instance, 14.5 KLO (IF). If my inbound course is 242º, can I break that course before D14.5 to join the final course 155º, or should I first reach 14.5 (overshoot then) and go catch the 155º?
Thank you vm

Good question. I was taught to anticipate turns and my Garmin 430 will anticipate turns so that you do not overfly the fix. But I don’t remember any specific guidance on the matter, so I looked it up. I found this in the AIM.

5−3−5. Airway or Route Course Changes
a. Pilots of aircraft are required to adhere to airways or routes
  being flown. Special attention must be given to this requirement
  during course changes. Each course change consists of variables that
  make the technique applicable in each case a matter only the pilot can
  resolve. Some variables which must be considered are turn radius, wind
  effect, airspeed, degree of turn, and cockpit instrumentation. An
  early turn, as illustrated below, is one method of adhering to airways
  or routes. The use of any available cockpit instrumentation, such as
  Distance Measuring Equipment, may be used by the pilot to lead the
  turn when making course changes. This is consistent with the intent of
  14 CFR Section 91.181, which requires pilots to operate along the
  centerline of an airway and along the direct course between
  navigational aids or fixes.
b. Turns which begin at or after fix passage may exceed airway or
  route boundaries. FIG 5−3−1 contains an example flight track depicting
  this, together with an example of an early turn.
c. Without such actions as leading a turn, aircraft operating in
  excess of 290 knots true air speed (TAS) can exceed the normal airway
  or route boundaries… Consequently, the FAA expects pilots to lead
  turns and take other actions they consider necessary during course
  changes to adhere as closely as possible to the airways or route being
  flown.

Good answer! I am not sure from the quote title whether that paragraph applies to terminal area procedures? Also the OP question mentions an EASA example, so FAA rules help for understanding, but EASA rules could differ.

The use of a circle around the waypoint symbol to indicate fly-over waypoints holds good for both RNAV and non-RNAV waypoints,
Earlier, a note on the chart using wording similar to "do not turn before" has been used to identify a flyover WPT.
Unless you see a circled WPT or an instruction note you treat all WPTs as fly-by points.
(Tip: ensure your speed is commensurate with the distance to touchdown, 'speed control', and smaller radius of turn.)


text:
Does Daylight Saving Time affect local airspace?
Only some countries use Daylight Saving Time. In Canada, some provinces use and don't use DST (such as Saskatchewan). However, in remote and local regions where this is not observed, what would happen?
What happens when areas that do not observe DST are controlled by areas that do. (Cleveland controls parts of Southern Ontario). How does this affect local airspace? Does it affect global airspace as well? What are the safety implications? How does it affect Air Traffic Controllers operating these different spaces?

@Zizouz212: Just pick one at random to accept then. Accepting one of the answers will make the question show up as 'answered satisfactorily' in the list of questions.

There is no such thing as "Daylight Saving Time" in the first place. Daytime is the same regardless of what your clock is set to show.

Aviation always uses UTC time. This is from the FAA's AIM, but other countries do the same for obvious reasons:

4-2-12. Time
a. FAA uses Coordinated Universal Time (UTC) for all operations. The
  word "local" or the time zone equivalent must be used to denote local
  when local time is given during radio and telephone communications.
  The term "Zulu" may be used to denote UTC.

This question covers in much more detail why UTC is used in preference to GMT, and this question explains why pilots need to know the time in the first place, at least under IFR (which applies to almost all airline flights).

Does daylight saving time affect local airspace?

Yes and No.
In aviation Flight plan information, weather, etc. all use Zulu Time (UTC). This provides a single time standard everyone in the world can use, and avoids misunderstandings based on various national rules for "Daylight Saving Time", "Summer Time", etc.
The airline / airport / travel agent / etc. are responsible for properly converting UTC times to the local timezone for passenger convenience.
Similarly active times for restricted airspace, TFRs, etc. are all published as UTC times to eliminate any possible ambiguity in this regard.

Not everything is based on UTC however: The (FAA-operated) tower at the field I fly from operates from 0700–2300 Local Time. When the United States messes with the clocks twice a year the UTC hours of operation/effective hours for the Class D airspace will change to remain aligned with the same "local time" operating period.

Air Traffic Control uses UTC or Zulu time which is the same, all the time, the whole world over.
Local only flights will use local time.

Yes it does, some Major international airports, notably SYD, have a curfew in place. Commercial passenger aircraft land between 0600 and 23:00 AEST or face sanctions. This in turn affects patterns in Dubai, Singapore, KL, LAX, DFW etc. This is a "must depart by" issue for some of these.
This is a politically motivated rule to appease voters in the flight path and is tied to "local" time. Aircraft arriving early are slowed and fly laps over suburbs, sharing the joy.

It does have an effect, look in St-Hubert, they have to work an hour longer in the summer!  That might be true anyway, but when reading times in the CFS there are always adjustments for the time change.  Essentially the airport (or parts thereof) opens and closes at the same local time but the zulu time changes.

TWR St-Hubert 118.4 352.5 (E) 1045-05Z‡ Apr-Oct; 1045-04Z Nov-Mar

You may think that doesn't happen in Sask., but look-

TWR Saskatoon 118.3 244.7 (E) 1200-0445Z Mon-Fri Mar 11-Nov 3,
  1245-0445Z Sat-Sun Mar 11-Nov 3, 1245-0445Z Nov 4-Mar 10

They also have to make adjustments to their times to accommodate traffic from other provinces.  They also work longer in the summer.
This is from an old CFS I found online, don't land and say I said they would be open.

In all air traffic control systems, all time is UTC (called "Zulu time"), which doesn't have daylight savings. Local time is only used when communicating with passengers (departure/arrival boards, announcements, boarding tickets, etc).

Some class D towers are only open part time based on local time. So yes, daylight savings time will impact when the tower and class D airspace is active.


text:
What is CAPT & PURS?
On the airbus A320 Evacuation Control Panel, what is CAPT & PURS. On some takeoffs it is selected and on other takeoffs it is not selected. What does this stand for and what is it? What does it do?
Here is a graphical representation:

It is a two position tumbler switch on the overhead panel to allow/disable the activation of the Evacuation alert from the cabin (explained here page 13). The view can also be found here for A320 and also here, with the following comments:

CAPT and PURS - the alert may be activated from either cockpit or cabin.
CAPT - the alert may be activated from cockpit only.

The A320 pictures seem containing the safety wire to keep the switch in the CAPT and PURS position, so it most likely another position is only selected for the reason.
The manual says Allows activation of an Evacuation alert from
the cockpit or cabin but from explanation above and the notation on the panel looks like the alert activation from the cocpit cannot be disabled by this switch.
This electric component is also called "toggle switch" and may be called "flush switch". These components were patented as early as in 1917.

I assume the abbreviations stand for "captain" and "purser?"

I thought you used "tumbler" by mistake, but I see that it is a valid use of the word. Maybe a Commonwealth English term? Is it a term that Airbus uses? I am used to using "toggle switch" instead.

The discussion of the history and nomenclature of toggle switches doesn't seem to be germane to the question


text:
How can I figure out the take-off minimum when multiple minimums are listed in the TPP?
For example:
If the published take-off minimum depicts:

Standard with minimum climb of 565' per NM to 2800, or 1000-3 with
  minimum climb of 370' per NM to 2800, or 1500-3 for climb in visual
  conditions.

How can I figure out which of the various minimums apply? For example, if the airport has 1200-6, which of the available take-off minima can be used? Could you explain in detail and why?

Related and maybe this one

(this answer is specific to the USA and FAA rules. Other countries may be the same, or similar, but I'm not familiar enough with them to guaruntee that this answer can apply outside of the USA)

To answer your question, we must first define what a "Standard Takeoff Minimum" is. Part 91.175 defines the standard wx conditions required for takeoff for commercial operators on an IFR flight plan:

§91.175 Takeoff and landing under IFR. (f) (2) (i) For aircraft, other than helicopters, having two engines or less—1 statute mile visibility. (ii) For aircraft having more than two engines— 1⁄2 statute mile visibility.

Notice that there are no cieling requirements, only visibility. One- and two-engine airplanes must have 1 mile of visibility, and airplanes with 3 or more engines require 1/2 mile visibility.

Non-Standard - Some runways have obstacles in the distance or, for other reasons, require the weather to be higher than "standard" in order to use that runway during low visibility operations. These non-standard" minimums are published in the TPP. These minimums might be visibility-limiting only, or they might require both a minimum visibility as well as a minimum ceiling.

Ok, so that gets us to your question, in which you have provided specific examples.

Question: If take-off minimum depicts "Standard with minimum climb of 565' per NM to 2800, or 1000-3 with minimum climb of 370' per NM to 2800, or 1500-3 for climb in visual conditions."

Standard with minimum climb of 565' per nm to 2800. This means that if your aircraft has the capability to climb at a rate of 565 feet per nm all the way to 2800 feet MSL, then you can use "standard takeoff minimums", which is 1 mile viz for 2 engines or less, or 1/2 mile for 3 engines or more, as long as you maintain that climb rate all the way to 2800'.
or 1000-3 with minimum climb of 370' per NM to 2800. If your aircraft cannot maintain the climb performance listed in #1, but you CAN maintain a 370'/nm rate of climb, then you can takeoff when the weather conditions are 1000-3.
or 1500-3 for climb in visual conditions. If you cannot maintain either of the two previous climb-out requirements, then you must takeoff in visual conditions and the weather observation must be a minimum of 1500-3

Question: If the airport has 1200-6, what take-off minimum can be used?

In this case, you have a visibility of 6 miles, which is higher than the minimum visibility requirements for all 3 items listed above.  But you also have a ceiling of 1200 feet, which would prevent you from using Item #3 (due to the minimum ceiling requirement of 1500).
If your aircraft has the performance capability to meet either Item #1 or #2, then you have legally met the minimums and can depart.

IMPORTANT Follow-On

The above information is only applicable to commercial flights operating under Parts 121, 125, 129, and 135 rules.  "Takeoff Minimums" do not apply to flights operating under Part 91 rules.  A part 91 flight can legally depart under 0-0 wx conditions. (although it is generally considered unsafe to do so, and several arguments have been made that it might be a violation of the rules to "operate safely"

It is important to note that minimums in §91.175 do not apply to most of us (Part 91 General Aviation). They do apply most —but not all— commercial operations like the airlines, freight,  and charter operations. Charted limitations on the other hand apply to everyone.


text:
How can we avoid running out of airport abbreviations?
As far as I know, airport name abbreviations are a 3 letter name. For example:

MEL for Melbourne/Australia airport
LAX for Los Angeles/USA airport
YYZ for Toronto/Canada airport

And that means all the options we have for naming are 26 * 26 * 26 = 17,576
Sometimes these match the IATA code, but not always. IATA codes are also 3-letter codes, but are not assigned to every airport as shown in this answer. But what about all of the other airports in the world? Will the world ever run out of "regular" airport codes?
A simple Google search about the number of the airports worldwide shows there are 17,678 airports. So are they going to extend airport codes to 4 letters or use numbers 0,1,2,...9?
Even by starting to use numbers they may reach a point where they run out of options. What is the approach to fix that ?

There are a lot of airports that use numeric codes, for example a popular airport near my home airport is 92C - Carter.

are you interested in IATA airport codes or ICAO airport codes?

@selectstriker2 not sure what you call these codes MEL, LAX, YYZ

To understand the difference between ICAO and IATA codes you can refer to this Wikipedia article.

They are not IATA airport identifiers, but IATA location identifiers. See this answer about railway stations having IATA identifiers too. So the problem is larger :-) Here is the list of these stations.

@mins and in a similar vein, ICAO codes are also not just assigned to airports.  ARTCCs, Weather Forecasting Offices, and some other aviation-related-but-not-airport functions all get ICAO codes as well.

This was marked as a duplicate, but I disagree.  The question is about "regular" airport codes, assigned to every airport in the world.  It is not about just IATA codes, which is what is being flagged as a duplicate.  I've slightly edited this question to make it more unique.

@Jimmy: As you mention other series, also food/meals.

There are multiple airport code systems. The one you seem to be asking about is the 3-letter IATA code, which is mostly assigned to airports with scheduled commercial service--and even some railway stations; they're nowhere near running out.
4-letter ICAO codes are assigned to far more airports, and we're nowhere near running out of those globally either, though since countries are allocated 1- or 2-letter prefixes, it's rather uneven.
Some airports have only a three- or four-letter national code; this is particularly common for heliports and private fields, and one or more of the letters is often a number so they aren't confused with IATA or ICAO codes.

Only major airports get all Letters.   Many of the smaller airports also incorporate numbers.
Here are some examples:
Osage City: 53K
Cle Elum: S93
De Vere: 2W1
And many, many, many more...
Once you add in letters and numbers, you'll find plenty of combinations.

Which, assuming 0 is not included, then becomes (26+9)3535 = 44100

0 is used, frequently for sea-plane areas, such as 0W0 (Seattle)

These are called Airport Identification Codes. The same ID code might exist in different countries, so it's not really possible to limit your 17,576 total combinations to all airports in the world. Additionally, some countries will use a combination of letters and numbers.
There are three main conventions for determining an Airport ID code:

Each country has their own national authority, which assigns IDs to airports in that country
ICAO - The ICAO organization has a process for each country to assign an ICAO ID to airports within that country.
IATA - The IATA organization has a process for each country to assign an IATA ID to airprots within that country.

National Authority:
Within each country, the national aviation authority assigns Airport IDs to all airports within its jurisdiction.  For example, in the USA, the FAA assigns either 3- or 4-letter codes to its airports.  There is more information on FAA-specific codes at this ASE page. The busier airports in the USA will get a 3-letter ID. These IDs will usually match the IATA code. And, within the 48-contiguous states, the 3-letter ID will be prefaced with the ICAO-designated letter "K".
International Civil Aviation Organization (ICAO).
This body assigns a prefix designator letter to geographical regions and (if necessary) a second letter to individual countries within that region.  Some examples: Northern European airports all start with E, and then the UK starts with G and Germany starts with D.  From there, the national authority selects the final two letters.  Thus, London Heathrow's ICAO Code is: EGLL.
International Air Transport Association (IATA)
IATA is not driven by the local authority, but rather by the Association itself, in cooperation with the member airlines.  IATA assigns a 3-letter code to airports which receive service by IATA-member airlines.
And to get even more complicated, the same airport might have three different ID Codes, one of each.

Related reading:
Can two airports have a different ICAO code but share the same IATA code?
When do we use IATA codes and when do we use ICAO codes?

as far as i know airport names abbreviation is a 3 letters name

There are multiple lists of airports that use abbreviations and none of them are completely comprehensive.
The major lists you will see for airports around the world will be IATA and ICAO.  The FAA has a separate list that is only for US airports.  The three-letter abbreviations you show are from the IATA list.
Many smaller airports do not get an IATA designation.  Checking a recent version of the data shows only a little over 9000 entries.  So unless the allocation rate is high (which seems unlikely), there's no immediate fear of running out.
See also: Can two airports have a different ICAO code but share the same IATA code?
and Where can I find a free list of ICAO and IATA airport identifiers?


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Why is the force that moves an airplane forward called THRUST and not PULL?
As we know, there are four forces acting upon an airplane. Thrust ($T$), Drag ($D$), Lift ($L$), and weight ($W$).
Since most propeller-driven airplanes over history have had the propeller mounted in the front, the power or forces generated were to pull, not to push. So why is it called thrust (the meaning of which is equivalent to push), and not pull??

How do you pull air?

Well, wouldn't that would be about like calling lift "suck"?  And we all know that lift doesn't suck.  Also, you'd have a problem explaining jet propulsion, and even more so rockets.

Thrust is the force reaction from accelerating a gas from here to over there.  If the element creating the thrust, a propeller, is in front of the engine, it's called a "tractor" configuration because it's pulling the engine, like a tractor (any normal front engine airplane).  If the prop is at the back, or more correctly, behind the engine, it's a pusher configuration (A Lake Buccaneer amphibian is a pusher even though the engine is amidships, not at the tail, because the prop is on the back of the engine itself).  Lots of other designs have put the prop all the way at the back.
In a jet engine, you don't really have an element pushing or pulling the engine itself because the thrust reaction is actually happening inside the engine, mostly in the area ahead the convergent nozzle of the tailpipe where the flow is forced to accelerate. So for jet engines it's just thrust.

After I checked again, the first design of airplane, that was the Wright flyer, was a canard rather then tail. The propeller is behind the wing to push the airplane. So, very possible it is the reason the name is called thrust, which literally means pushing. While the resistance suffered by the airplane is called drag.

Pushing and pulling should be considered in relation to the air, not the engine. For many means of propulsion, there simply is little in the way of any pulling going on. Best example of this is a rocket, all force comes from the push out the back. With jet engines, the inlet often creates drag, it does not suction the aircraft forward normally. Propellers are a bit more tricky in this regard when you examine the math, as the lack of mass addition or significant heating makes the "inlet" and "outlet" conditions conserve a lot better, but it is still the aircraft acting on the air, not the other way around, and the air that has been acted on normally ends up behind, not in front of the aircraft.

Or, is it has relation to the previously using propeller, ship?

@AirCraftLover I don't understand the sentence you just wrote, but I'm guessing you meant to write: "Or, is it related to an older vehicle that has a propeller—a ship?"

If you think about it, suction is not pulling air, it is the pressure of the atmosphere pushing air into a place where you've created a partial vacuum.


text:
From a safety standpoint, which is better: voice aural alerts or tone aural alerts?
Most modern airplanes have some sort of crew alerting system. These alerts are generally conveyed through display messages, lights, and aural alerts. Aural alerts can either be tones or voices. For example, on many Boeing airplanes, most caution alerts are indicated by a beeper and most warning alerts are indicated by a siren. However, on most McDonnell Douglas airplanes, most alerts are indicated by a voice, such as in, "engine one fire," "landing gear," and "stabilizer motion." Most airplanes will have a mix of both, but generally "stick" to one form of aural alerts over the other. For example, most Airbus airplanes have tone aural alerts, but some specific conditions, such as a stall, will trigger a voice aural alert. Overall, which is better? Can voice alerts be more distracting, or are tones not specific enough to orient the crew to an abnormal condition?

In episode 94, Jamie Hyneman thought the simulator was saying, Warning. Don't Think. ... It was saying, Don't Sink. - He grew up in Michigan and has a degree in Russian linguistics, and not even he could understand the thing.

Older aircraft designed before the B787 typically had the variability in design that you refer to. When the B787 started development, the FAA issued special conditions that required Boeing to address issues the FAA had with the alerting.
The FAA was concerned about proper prioritization of alerts and reducing ambiguity. Eventually, the Special Conditions resulted in a revision to the regulations, § 25.1322. That was followed with a new Advisory Circular AC 25.1322-1 - Flightcrew Alerting.
Generally, voice alerts are where a specific response is required immediately, think TAWS or TCAS. The issue with voice alerts is that they can not overlap. Distinct audio alerts can exist simultaneously - e.g., overspeed with a "pull-up" ground proximity alert.
I suggest a read through of the AC to get a better understanding of the current design guidelines.

"The issue with voice alerts is that they can not overlap." But you can easily time interleave them. “BANK ANGLE, BANK ANGLE; PRESSURE, PRESSURE; BANK ANGLE, BANK ANGLE; PRESSURE, PRESSURE”. Learning what the different sounds mean is already difficult enough, if you have to learn all combinations of them it’s almost impossible.

@Michael As someone who spent 25 years developing avionics, I can say it is definitely not 'easy to interleave' the messages. Every alert is assigned a priority in the system. TAWS, Windshear, and TCAS have different responses and thus the voice alerts to identify the action to be taken. Once an alert starts it must finish (though not necessarily repeat). A higher priority alert can follow, but the lower alert won't resume until the higher priority is no longer present. The FAA is very concerned with misleading alerts caused by alerts run together. So allowable combinations are very limited.

This is only my personal experience from my initial-course CRJ simulator training, but I found that voice alerts tended to be more effective at distracting you or pulling you out of mental fog.
For example, with the Rockwell Collins Pro-line avionics suite, the tones and voice calls are distinctive to that system. The landing gear warning horn is a dual frequency continuous tone that (to me at least) sounds exactly like the old TV test pattern sound from the days of broadcast TV, when they shut down at 2AM each night.
I found that if pushed to the edge of mental saturation by abnormal and emergency tasks, (usually while nearing burnout late in a 4 hour session - instructors can be evil), if that tone came on during an extreme workload situation it would be very easy for it to just become part of the background noise if it wasn't noticed right away, whereas the "TOO LOW, GEAR, TOO LOW, GEAR" voice alert from the Ground Proximity Warning System, that would come on when you got lower, had a much better effect at bringing you out of the fog.
So if I had to choose one or the other, from an effectiveness perspective I'd go with voice alert, but there are limitations there as well.


text:
Do all 767-200s have speedbrakes that deploy automatically upon touchdown?
It may just be a coincidence, but I found two videos that involve the 767-200's speedbrakes not deploying automatically upon touchdown (which should occur if they are armed).
The first video shows the speedbrake lever move less than an inch upon touchdown, then the pilot flying extends them all the way.
The second video shows the speedbrakes not deploying at all for at least 5 seconds after touchdown, then they extend in incrementes until extending fully a few seconds after that.
Does the 767-200 have some sort of "quirk" when it comes to its speedbrakes deploying, do some 767s not have an "armed" function, or are these just coincidences where they failed to automatically deploy?
P.S. I also have this footage of the 757's first flight where the speedbrakes clearly do not deploy automatically, either.

All B757 and B767 aircraft have speedbrakes that can be deployed automatically. But you need to manually arm them first. You also need weight on wheels(truck tilt) and both thrust levers at idle, before they automatically deploy. Moving the reverse levers to the idle reverse detent will also raise the speedbrakes, even if they are not armed.
On some landings they will start to deploy and then stop if there is not enough weight on wheels, due to a little bounce or float. The flying pilot will then usually manually deploy them.
In your first video, it looks like the speedbrake was not armed. Your second video looks pretty normal, but the landing was so smooth, it’s hard to tell when the wheels were on the ground. It may have been a manual deployment, or deployment with the thrust reversers. The third B757 video looks like late deployment, so it may have been manual as well.

Thanks, @Mike Sowsun. But the speedbrakes only deploy after 4 seconds once even the nose gear is on the ground in the third (757) video. Did they fail to deploy or were they probably simply not armed?

Thanks guys, meant to say “idle reverse detent”, so I edited my answer. On the 757 video, it does look like a late deployment, so I edited that part too.

Yup, thanks for being a proofreader. :)

Is this still true for the small number of 767s that were produced with manual (three-person) cockpits?


text:
Why are beacon lights used in conjunction with strobe and navigation lights in flight?
Strobe lights are much brighter and easier to see than beacon lights while in flight—it’s their function. Beacon lights are useful on the ground to indicate engines are running, but what is their use in the air? Strobe and navigation lights do a better job at establishing visibility and orientating other aircraft, while beacon lights are harder to see.

Welcome to Av.SE!

Beacons pre-date both LED and xenon strobes.  A number of modern aircraft have been certified without beacons because their strobe systems meet the angle-of-view requirements both vertically and horizontally without needing a beacon.  Cirrus aircraft, for instance, do not have beacons.
But the FAA position for anti-collision lighting has been "whatever anti-collision lighting the airplane was certified with, has to work and has to be on at night" unless there's a reason (such as precipitation blinding) that it shouldn't be.  So technically a plane cannot be flown if the beacon is inop even if the strobes meet the certification requirement, or even if you aircraft was certificated prior to the lighting rule changes.  A Legal Interpretation Letter from the FAA formalizes that even if your aircraft doesn't require them, if you do have them, they do have to work.  AOPA Legal did a write up on this a few years ago as a result of the letter.  AOPA Article
If your aircraft came with a factory beacon, it is very likely because either:

strobes were either added after-market, or;
the factory strobes did not meet the angle of visibility requirements at the time of certification.

Do Cirruses only have wingtip lights, no tail light? I know smaller planes that do, didn't know if that was true for Cirri.

@tedder42 Correct.  The wingtip lights units protrude from the wingtip enough that they meet the angular visibility requirements for certification.


text:
Why are taxiway and runway centerline lights off center?
The centerline lights for runways and taxiways are to the side of the centerline itself by a few inches. This, of course, is not a problem, but why are the lights positioned that way?

(Image: Hartsfield-Jackson Atlanta International Airport by John Murphy)

great photo of the one and only ATL there, BTW !

The painted line and the lights do not align for a very practical reason: it would be a major pain in the behind to paint over the lights, as each one of them would need to be masked for the painting, and have the masking removed afterwards.
Another reason, as pointed out in the comments by JScarry & Ralph J, is that placing the lights slightly to the side will keep the nosewheel from running over the lights. While the lights are designed to stand this, the constant thumping will put unnecessary strain on the landing gear, crew and the passengers.

Another reason is that the correct taxi and landing procedure is to have the nosewheel on the centerline. It probably won’t be noticeable in airliners but I can definitely hear and notice when I’m off center and running over the lights. It probably adds wear and tear to the lights if a big jet taxis over them.

Oh, it's noticeable in a 737, too - the nosewheel is just behind the pilots, and hitting all the lights isn't considered ideal for us either. (For a 747, probably not as much.)

The lights themselves are designed to  survive constant beating from big planes, but as stated, for planes, pilots and passengers running over them is not optimal. For the landing gear in particular it causes unnecessary wear.

@RalphJ I remember reading a comment by Terry saying you can even feel going over the lights in a 747.

@RalphJ as a Terry Fan I would like to add that running over the lights was his time killing hobby, even a competition with his co-pilot.


text:
I built my 1500 hours outside US. Will airlines in US still consider them as valid hours?
All of my flying licenses are FAA-based, ranging from PPL to IFR, CPL, and even a frozen ATPL. I acquired these licenses in Nepal, where I accumulated my 1500 flight hours as a first officer. In Nepal, the requirement is 250 hours and a frozen ATPL to serve as a first officer, allowing me to bypass becoming a CFI and gain experience in jet aircraft.
I am soon planning to come to the USA to obtain my unrestricted ATP license. However, I no longer wish to work in Nepal and would prefer to pursue opportunities in the USA. My logbooks are meticulously maintained, and I possess a valid work permit that allows me to work legally in the US.
Nevertheless, I have heard that American airlines may still require pilots to accumulate flight hours within their airspace, and they might not readily consider the hours earned in a different country under a different license.
I find myself in a state of confusion as I'm unsure whether this information is true or merely a rumor.

More the other way 'round, you would think lol. A pilot with 1500 hours flying between Oakland, LAX and Chicago and pushing "autoland"... would that really count as experience in a highly mountainous country full of interesting airports?

@Harper-ReinstateMonica The issue is fraud. It's not unheard of to create a position and flight hours with a bribe in some countries. Further, after some high-profile fatal crashes with pilots hiding information, carriers are required to enter and review pilot documents in the FAA Pilot Records Database.

The FAA regulations on Aeronautical Experience requirements do not specify that the requirements have to be met while flying in the US, or even with an FAA pilot certificate. As long as your logbooks pass muster and you meet the requirements, you qualify.


text:
What is this military aircraft I saw near Catalina island?
I was wondering if anyone could help me identify this plane. I have researched military id guides, image searched, and asked several naval and Air Force friends with no result.

Location: off shore 20 miles south of Catalina island, California
Date/time: 5pm June 29, 2023

Plane flew lowish (1000ft) over our boat, turned around and did it again. It is jet powered, fuselage mount I think.
Closest match I can find is F86 Sabre or Grumman F9 cougar. Don't think it's either one though.
It appeared all grey with no markings that I could discern.  The images show a camouflage pattern but I think that's shadows?

It looks like a Hawker Hunter (don't know if it still around and if it has flown in the US though)

Thank you so much Sophit!  You were correct!

You're welcome  I didn't elaborate an answer because I wasn't sure if it were still airworthy but @jamiec gave a super answer

If you get Smithsonian Channel, an episode of Air Warriors about the Hawker Hunter is playing periodically this week.

Thanks for the tip!

Almost certainly a Hawker Hunter, in fact the image on wiki is almost exactly the same aspect as your first picture:

One has certainly been spotted near Catalina Island before:

Here is a photo of the exact aircraft:

N330AX flight on 29th June at around 3:30pm Pacific Time did indeed take it near Catalina Island - and I will screenshot flightradar below because it wont be available forever to non paid members:

Well, I'm certainly a believer.  You're absolutely right, that's the plane.  Amazing you pulled all that together so quickly.  Thank you! I had been looking a lot and it was bothering me.  BTW - we were 20 miles east of Catalina, so it all matches up.  Have to say it was super fun to watch this plane fly low. Was super loud and powerful.

From two blurry photos to complete identification of not only the exact aircraft but the exact mission, in about an hour. There are no secrets any more.

@A.I.Breveleri Mostly luck in this case, and the fact there really cannot be many hawker hunters in west coast USA

Would this be for something like the USAF Heritage Flight Conference at Davis Monthan? The Navy equivalent?

This aircraft is one of a fleet of Hunters operated by ATAC, a PMC based at Point Mugu.

Oh cool @ColinPickard good to know.

The 2nd photo seem to show (to me at least) something large attached to the bottom of the plane at the back. Does the Hawker Hunter have a standard attach point there (for say a bomb, a fuel drop-tank or some other payload)? Or perhaps could that be under a wing but viewed at a strange angle?

@uhoh, I think that's a tank mounted on the near wing.  The first photo certainly shows wing tanks (or wing somethings). I don't see how one can distinguish a wing-mounted tank from a fuselage-mounted one in the second photo, other than that you can only see one.  But you can't see the far wing at all, and if the two photos show the same plane, then the far wing's tank must be obscured by the fuselage in the second photo.


text:
Why are British Airways wings so dirty?
I often fly through Heathrow, and British Airways airplanes, particularly their 777s, always look so dirty. For example, there is often a kind of dark oily stain on the wings and flight surface. So - why are BA planes so dirty compared to the others? And what exactly is this dirty anyway?

(image from this YouTube video)

Related: Do airlines ever clean their planes?

What evidence do you have that this is a significant issue on BA aircraft only? Or is it a 777 issue? Or a BA 777 only issue? One image from one video seems to lead to an unsubstantiated claim of "always". Maybe next time you're through LHR, take a pic of every plane parked at every gate you can get to and do a count to see if BA is really the dirtiest airline. As it stands, this seems to be unsubstantiated speculation...

@FreeMan I fly through Heathrow at least eight times a year, and have been doing so for the past 20 years. As I have said in the post, I observe dirt on BA planes, particularly on their 777 fleet, that stand out compared to other airlines. I attached an image that is a representative sample of my observations. If you choose not to believe me, that is fine - I am simply looking for a plausible explanation.

It's probably because they want them to match the dirty interior. BA is not known for the quality of their housekeeping, and I have flown them in the US, the EU, the Middle East and Africa.

@JuanJimenez It's a shame that a flag carrier, of a country known for its aviation heritage, should present itself in such an unappealing way :(

To match their 'tricks'  - https://en.wikipedia.org/wiki/Dirty_Tricks_(scandal)


text:
What exactly are these two cones on top of this Sikorsky S-61 helicopter?
Haven't notice this thing on the Sikorsky, what is it?

Another picture from another angle:

These are engine inlet screens, to prevent FOD ingestion. See this question for a very similar device:

Thats what I thought but wasn't sure. Do you want to post it as an answer so I can accept it or should I delete this post?

These are engine inlet screens, to prevent FOD ingestion. See this question for a very similar device.

These seem to have an additional trick up their sleeves, in that they are not just screens but also deflectors.

yup, that's what initially caught my attention, that's a good idea so that the FOD doesn't stay fixed to the screen by the suction force.


text:
What is this display for on the Embraer E175?
Caught my attention, never had I seen a display in this kind of overhead position in an aircraft. What is it for?

Source of the photo?

@RalphJ and @ Michael actual aircraft, it seems not all ERJs have that, an ERJ pilot out there should probably know. I found it on google images, can't find again the same picture link, found another one https://c2.staticflickr.com/6/5139/5529861633_4822a964c5_b.jpg

still the coolest yokes I've seen

That is the screen for the Cockpit Door Surveillance System (CDSS) which can show a live video feed of the area in front of the cockpit door. This allows pilots to decide whether it's safe to open the door when someone requests access to the cockpit, e.g. the other pilot returning from the lavatory.
I couldn't find the exact model used on this Embraer, but here is a similar system:

Three cameras are located in the passenger cabin ceiling panels. One camera is installed immediately in front of the cockpit door. Once camera is above the forward passenger door and one is above the forward service door. The combination of camera locations provides full visibility of the forward passenger cabin area from in front of the passenger door, right across to the service door, including the galley area. When the pilot is checking the system to identify who is requesting access to the cockpit, it is also possible to see that there is nobody else hiding behind the toilet bulkhead or in the galley. It is also possible to identify that the person outside the galley door is not under threat or duress.

(AD Aerospace: Cockpit Door Surveillance System (CDSS))
The newer E-Jets E2 use a slightly different model:

(image source)

I was considering a smart-mouth answer of how its a rearview mirror (cos its in a similar relative position to one in a car)  but dang it basically is !

@Criggie GMTA I was going to say "it's that extra rear view mirror so you can see what the kids are up to in the back seat of the Sienna .."